why assert(str!=NULL) does not return an error?

July 24, 2022

Why did this not print an assert error?

This is my code:

#include<assert.h>
#include<stdio.h>

int main(){
    int n =12;
    char str[50] = "";
    assert(n>=10);
    printf("output :%d\n",n);
    assert(str!=NULL);
    printf("output :%s\n",str);
}

>Solution :

char str[50] = "";

This makes str into an array of 50 chars, and initialises the memory to all zeroes (first byte explicitly from "" and rest implicitly, because C does not support partial initialisation of arrays or structs).

assert(str!=NULL);

When used in an expression, array is treated as pointer to its first element. The first element of the array very much has an address, so it is not NULL.

If you want to test if the first element of the array is 0, meaning empty string you need

assert(str[0] != '\0');

You could compare to 0, or just say assert(*str);, but comparing to character literal '\0' makes it explicit to the reader of the code, that you are probably testing for string-terminating zero byte, not some other kind of zero, even if for the C compiler they’re all the same.